is 54 e Haber process is the principal method for fixing nitrogen (converting N2

Question

is 54 e Haber process is the principal method for fixing nitrogen (converting N2 to nitrogen compounds). N2(g) + 3H2(g)- 2 NH3(g) Assume that the reactant gases are completely con- verted to NH3(g) and that the gases behave ideally (a) What volume of NH3(g) can be produced from 152 L N2(g) and 313 L of H2(g) if the gases are mea- sured at 315 °C and 5.25 atm? (b) What volume of NH3(g), measured at 25 °C and 727 mmHg, can be produced from 152 L N2(g) and 313 L H2(g), measured at 315°C and 5.25 atm?

Explanation / Answer

Answer:

Given that N2 (g) +3 H2 (g)----------->2 NH3 (g)

From above equation, we observed that 1 mole of N2 reacts with 3 moles of H2 to form 2 moles of NH3.

a volume of N2 and H2 are given-

a volume of N2 (v1) = 152 l

the volume of H2 (v2) = 313 l

temperature = 315o C = 588 K, pressure = 5.25 atm

a) So, we can calculate how many moles of N2 and H2 gases are reacting with each other using an ideal gas equation, after that we can find out the volume of NH3 using stoichiometric coefficient balance.

So

moles of N2 (n1) = P1V1 / RT = 5.25 atm * 152 l / (0.082 L atm K1 mol1 * 588 K)

n1 = 16.55 moles of N2

moles of H2 (n2) = P2V2 /RT = 5.25 * 313 l /  (0.082 L atm K1 mol1 * 588 K)

n2 = 34.08 moles of H2

Since limiting reagent in above reaction is N2 gas, and we observed that from 1 mole of N2 gas there are 3 moles of NH3 is forming so,

if we used 16.55 moles of N2 gas then

moles of NH3 will form = 3 * 16.55  

moles of NH3 form = 49.85 moles of NH3

For the volume of NH3 use an ideal gas equation.

V = n3RT/ P = 49.85 * .082 *588 L atm K1 mol1 / 5.25 atm

V = 457.822 l -- answer

b) At standard condition

volume of ideal gas = 22.4 l / mol

So, we can calculate moles of N2 and H2 gas by dividing the volume of gases by standard volume

moles of N2 = 152 / 22.4 = 6.785

moles of H2 = 313l / 22.4 = 13.97

Again, limiting reagent is N2 so,

moles of NH3 will form = 3 * 6.785 = 20.355

use an ideal gas equation to find out volume of NH3

V = n3RT / P  

given T = 273 K

P = 727 mmhg = 0.956 atm

R = 0.082 L atm K1 mol1

So, V = 20.355 * 0.082 L atm K1 mol1 * 273 K / .956 atm

V = 476.63 l volume of NH3 will form -- answer

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